Problem 64 - Calculus for Biology and Medicine

Question

Use the determinant to determine whether $$B=\left[\begin{array}{rr} 1 & 1 \\ -1 & 1 \end{array}\right]$$ is invertible. If it is invertible, compute its inverse. In either case. solve \(B X=\mathbf{0}\).

Step-by-Step Solution

Answer

The matrix \(B\) is invertible, and its inverse is \(B^{-1} = \begin{bmatrix}\frac{1}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{1}{2}\end{bmatrix}\). The solution to \(BX=\mathbf{0}\) is \(X=\mathbf{0}\).

Step 1: Understanding Invertibility

A matrix is invertible if and only if its determinant is non-zero. For a 2x2 matrix \(B=\begin{bmatrix} a & b \ c & d \end{bmatrix}\), the determinant is calculated as \(\text{det}(B) = ad - bc\).

Step 2: Calculate the Determinant

For the matrix \(B=\begin{bmatrix} 1 & 1 \ -1 & 1 \end{bmatrix}\), calculate the determinant using \(\text{det}(B) = (1)(1) - (-1)(1) = 1 + 1 = 2\). Thus, \(\text{det}(B)=2\), which means \(B\) is invertible since the determinant is non-zero.

Step 3: Inverse of the Matrix

The inverse of a 2x2 matrix \(B=\begin{bmatrix} a & b \ c & d \end{bmatrix}\) is given by \(B^{-1} = \frac{1}{\text{det}(B)} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}\). Substitute the values from matrix \(B\): \(B^{-1} = \frac{1}{2} \begin{bmatrix} 1 & -1 \ 1 & 1 \end{bmatrix} = \begin{bmatrix} \frac{1}{2} & -\frac{1}{2} \ \frac{1}{2} & \frac{1}{2} \end{bmatrix}\).

Step 4: Solve the Equation BX = 0

Since \(B\) is invertible, the only solution to the equation \(BX = \mathbf{0}\) is the trivial solution \(X=\mathbf{0}\). This is because if \(B^{-1}\) exists, then \(X = B^{-1} \cdot \mathbf{0} = \mathbf{0}\).

Key Concepts

Invertibility

To determine if a matrix is invertible, we look at its determinant. The determinant is essentially a special number that can tell us a lot about the nature of the matrix. If the determinant is not zero, then the matrix is invertible. This means there is another matrix that, when multiplied with it, will result in the identity matrix, much like how a number and its reciprocal multiply to 1.
For any 2x2 matrix, represented as:

• \( B = \begin{bmatrix} a & b \ c & d \end{bmatrix} \)

The determinant can be calculated using the formula:

• \( \text{det}(B) = ad - bc \)

In simple terms, if \( \text{det}(B) eq 0 \), we have a sure sign that the matrix is invertible. So, for the given matrix \( B \), we calculated its determinant to be 2. This is why \( B \) is invertible. It means there exists some inverse matrix for \( B \).

Inverse Matrix

Finding an inverse matrix is a crucial step when dealing with invertible matrices. The inverse matrix essentially behaves like a number's reciprocal. For example, multiplying a number by its reciprocal gives you 1. Similarly, multiplying a matrix by its inverse results in the identity matrix. This identity matrix acts like the number 1 in matrix multiplication.
For a 2x2 matrix \( B = \begin{bmatrix} a & b \ c & d \end{bmatrix} \), if \( \text{det}(B) eq 0 \), then the inverse is:

• \( B^{-1} = \frac{1}{\text{det}(B)} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \)

To keep things simple, let's apply this to the given matrix:

• \( B = \begin{bmatrix} 1 & 1 \ -1 & 1 \end{bmatrix} \)

Its inverse is computed to be:

• \( B^{-1} = \begin{bmatrix} \frac{1}{2} & -\frac{1}{2} \ \frac{1}{2} & \frac{1}{2} \end{bmatrix} \)

The inverse allows you to "undo" the effect of the matrix. If you have a matrix equation, you can solve it by multiplying by the inverse. Just like dividing both sides of a simple algebraic equation by a number, you can "divide" by a matrix by multiplying by its inverse.

Linear Equations

Solving linear equations using matrices is a powerful method in algebra. When we talk about linear equations in the context of matrices, we often speak of the equation \( BX = \mathbf{0} \), where \( B \) is a matrix and \( X \) is a vector of variables. If \( B \) is invertible, this equation has a unique solution known as the trivial solution.
In the specific case where \( B \) is invertible and we're solving \( BX = \mathbf{0} \), the only solution is \( X = \mathbf{0} \). Why? Because if \( B \) is invertible, multiplying both sides of the equation by \( B^{-1} \) (the inverse of \( B \)) results in:

• \( X = B^{-1} \cdot \mathbf{0} = \mathbf{0} \)

This simplicity is one of the reasons understanding invertibility and finding inverse matrices is so important. It tells us if the system of equations represented by the matrix equation has no solutions or exactly one solution.
Learning to solve these equations helps in understanding a broad array of applications, from engineering problems to economic models, where systems of linear equations describe complex dynamics in a simpler abstraction.